Submission #830398

#	Time	Username	Problem	Language	Result	Execution time	Memory
830398		GrindMachine	Peru (RMI20_peru)	C++17	100 / 100	360 ms	78916 KiB

peru

// Om Namah Shivaya

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x, y) ((x + y - 1) / (y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i, n) for(int i = 0; i < n; ++i)
#define rep1(i, n) for(int i = 1; i <= n; ++i)
#define rev(i, s, e) for(int i = s; i >= e; --i)
#define trav(i, a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a, b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a, b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*



*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

#include "peru.h"

int solve(int n, int k, int* v){
    vector<ll> a(n+5);
    rep(i,n) a[i+1] = v[i];

    vector<ll> dp(n+5,inf2);
    dp[0] = 0;

    deque<ll> dq;
    multiset<ll> ms;

    rep1(i,n){
        while(!dq.empty() and a[i] > a[dq.back()]){
            ll j1 = dq.back();
            dq.pop_back();
            if(!dq.empty()){
                ll j2 = dq.back();
                ms.erase(ms.find(a[j1]+dp[j2]));
            }
        }

        while(!dq.empty() and dq.front() < i-k){
            ll j2 = dq.front();
            dq.pop_front();
            if(!dq.empty()){
                ll j1 = dq.front();
                ms.erase(ms.find(a[j1]+dp[j2]));
            }
        }

        if(!dq.empty()){
            ll j = dq.back();
            ms.insert(a[i]+dp[j]);
        }

        dq.pb(i);

        if(!ms.empty()){
            dp[i] = *ms.begin();
        }

        amin(dp[i],dp[max(i-k,0)]+a[dq[0]]);
    }

    // rep1(i,n){
    //     cout << dp[i] << " ";
    // }
    // cout << endl;

    ll ans = 0;
    rep1(i,n){
        ans = ans*23 + dp[i];
        ans %= MOD;
    }

    return ans;
}

• Subtask 1: 18 / 18
• Subtask 2: 31 / 31
• Subtask 3: 51 / 51